Problem 84 - Chemistry: Principles and Reactions

Question

Consider the following reaction at \(25^{\circ} \mathrm{C}\). $$ \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{Br}_{2}(l) $$ If \(\left[\mathrm{H}^{+}\right]\) is adjusted by adding a buffer that is \(0.100 M\) in sodium acetate and \(0.100 \mathrm{M}\) in acetic acid, the pressure of oxygen gas is \(1.00 \mathrm{~atm}\), and the bromide concentration is \(0.100 \mathrm{M}\), what is the calculated cell voltage? ( \(K_{\mathrm{a}}\) acetic acid \(\left.=1.8 \times 10^{-5} .\right)\).

Step-by-Step Solution

Answer

Answer: The cell voltage of the given reaction is approximately 0.36 V.

Step 1: Calculate Proton Concentration

We need to calculate the \(\left[\mathrm{H}^{+}\right]\) using the buffer system. The buffer system is composed of sodium acetate and acetic acid. The dissociation reaction for acetic acid is given by: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}^{+}$$ Using the given concentrations and the \(K_{\mathrm{a}}\) value, we can apply the Henderson-Hasselbalch equation to find the proton concentration: $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log{\frac{\left[\mathrm{CH}_{3}\mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3}\mathrm{COOH}\right]}}$$ Plug in the given values: $$\mathrm{pH} = -\log{(1.8 \times 10^{-5})} + \log{\frac{0.100}{0.100}} = -\log{(1.8 \times 10^{-5})}$$ Calculate the value of \(\mathrm{pH}\): $$\mathrm{pH} \approx 4.74$$ Now, we can calculate \(\left[\mathrm{H}^{+}\right]\): $$\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}} = 10^{-4.74} \approx 1.84 \times 10^{-5}\,\mathrm{M}$$

Step 2: Calculate Equilibrium Constant

Using the standard reduction potentials for the half-reactions, we can calculate the standard cell potential and equilibrium constant for the overall reaction at \(25^{\circ}\mathrm{C}\). The half-reactions are: $$\mathrm{O}_{2}(g) + 4 \ \mathrm{H}^{+}(aq) + 4 \ \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2}\mathrm{O}(l) \quad \quad E^{\circ} = 1.23\,\mathrm{V}$$ $$2 \mathrm{Br}^{-}(aq) \longrightarrow \mathrm{Br}_{2}(l) + 2 \ \mathrm{e}^{-} \quad \quad E^{\circ} = 1.09\,\mathrm{V}$$ The standard cell potential is: $$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ} = (1.23 - 1.09)\,\mathrm{V} = 0.14\,\mathrm{V}$$ Now, we can calculate the equilibrium constant \(K\) using the Nernst equation at \(25^{\circ}\mathrm{C}\): $$\Delta G_{\mathrm{cell}}^{\circ} = -nFE^{\circ} = -RT \ln{K}$$ Rearrange to solve for \(K\): $$K = \mathrm{e}^{\frac{-\Delta G_{\mathrm{cell}}^{\circ}}{RT}}$$ At \(25^{\circ}\mathrm{C}\), we have \(R = 8.314\,\mathrm{J\cdot mol^{-1} \cdot K^{-1}}\) and \(T = 298\,\mathrm{K}\). Keep in mind that the balanced overall reaction has \(n=4\) moles of electrons transferred. Plugging in the values, we can find \(K\): $$K = \mathrm{e}^{\frac{-(4)(96485\,\mathrm{C \cdot mol^{-1}})(0.14\,\mathrm{V})}{(8.314\,\mathrm{J \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}} \approx 36.1$$

Step 3: Calculate Cell Voltage

Finally, we can use the Nernst equation to calculate the cell voltage: $$E_{\mathrm{cell}} = E_{\mathrm{cell}}^{\circ} - \frac{RT}{nF} \ln{Q}$$ The reaction quotient \(Q\) is given by: $$Q = \frac{\left[\mathrm{H}_{2}\mathrm{O}\right]^2\left[\mathrm{Br}_{2}\right]^2}{\left[\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^4\left[\mathrm{Br}^{-}\right]^4}$$ Since the concentrations of \(\mathrm{H}_{2}\mathrm{O}\) and \(\mathrm{Br}_{2}\) are quite high as products, the reaction quotient simplifies to: $$Q \approx \frac{1}{\left[\mathrm{O}_{2}\right]\left[\mathrm{H}^{+}\right]^4\left[\mathrm{Br}^{-}\right]^4} = \frac{1}{(1.00\,\mathrm{atm})(1.84 \times 10^{-5}\,\mathrm{M})^4(0.100\,\mathrm{M})^4}$$ Replace the pressure of oxygen gas with its concentration using the ideal gas law: $$\left[\mathrm{O}_{2}\right] = \frac{n}{V} = \frac{P}{RT} = \frac{(1.00\,\mathrm{atm})(0.0821\,\mathrm{L \cdot atm \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}{1\,\mathrm{L}} \approx 0.0245\,\mathrm{M}$$ Insert the new value of \(\left[\mathrm{O}_{2}\right]\) into the reaction quotient: $$Q \approx \frac{1}{(0.0245\,\mathrm{M})(1.84 \times 10^{-5}\,\mathrm{M})^4(0.100\,\mathrm{M})^4} \approx 5.56 \times 10^{-12}$$ Now, we can evaluate the Nernst equation: $$E_{\mathrm{cell}} = 0.14\,\mathrm{V} - \frac{(8.314\,\mathrm{J \cdot mol^{-1} \cdot K^{-1}})(298\,\mathrm{K})}{(4)(96485\,\mathrm{C \cdot mol^{-1}})} \ln{(5.56 \times 10^{-12})} \approx 0.36\,\mathrm{V}$$ Thus, the calculated cell voltage is approximately \(0.36\,\mathrm{V}\).

Key Concepts

Nernst Equation

The Nernst Equation is a powerful tool in electrochemistry that allows us to calculate the voltage (or cell potential) of an electrochemical cell under non-standard conditions. It extends the concept of standard electrode potential by considering the effect of concentration on cell potential.
The basic Nernst Equation is given by:\[E_{cell} = E_{cell}^{\circ} - \frac{RT}{nF} \ln{Q} \]
where:

• \(E_{cell}\) = cell potential under non-standard conditions
• \(E_{cell}^{\circ}\) = standard cell potential
• \(R\) = universal gas constant
• \(T\) = temperature in Kelvin
• \(n\) = number of moles of electrons transferred
• \(F\) = Faraday constant
• \(Q\) = reaction quotient

The Nernst Equation shows how the cell potential can be adjusted for changes in ion concentration. As seen in the exercise, applying this equation involves calculating the reaction quotient, \(Q\), and then using it to adjust the standard potential to the specific conditions of the reaction.

Buffer Solutions

Buffer solutions resist changes in pH when small amounts of acid or base are added. They are crucial in maintaining a stable pH in various chemical and biological systems.
In the exercise, a buffer consisting of 0.100 M sodium acetate and 0.100 M acetic acid was used to control the concentration of hydrogen ions, \([\mathrm{H}^+]\). This is an example of an acetic acid-acetate buffer, which is commonly used in laboratories.

To determine the pH of this buffer, we use the Henderson-Hasselbalch equation:\[\text{pH} = \text{pK}_a + \log{\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)} \]
where:

• \([\text{A}^-]\) is the concentration of the acetate ion
• \([\text{HA}]\) is the concentration of acetic acid

This equation allows us to calculate the pH based on the concentrations of the weak acid and its conjugate base. The calculated pH enables us to find \([\mathrm{H}^+]\), which is used in the Nernst Equation to ascertain the cell potential.

Standard Reduction Potentials

Standard reduction potentials are a measure of the tendency of a chemical species to gain electrons and thereby be reduced. These values are determined under standard conditions (1 M concentration for solutions, 1 atm pressure for gases, and a pure solid or liquid state for other substances at 25°C).
In this exercise, we used standard reduction potentials to calculate the standard cell potential \(E_{cell}^{\circ}\). The involved half-reactions were:

• Reduction of oxygen: \(\mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4\mathrm{e}^- \longrightarrow 2\mathrm{H}_2\mathrm{O}(l),\ E^{\circ} = 1.23 \ \mathrm{V}\)
• Oxidation of bromide: \(2\mathrm{Br}^-(aq) \longrightarrow \mathrm{Br}_2(l) + 2\mathrm{e}^-,\ E^{\circ} = -1.09 \ \mathrm{V}\)

To find \(E_{cell}^{\circ}\), we use:\[E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}\]The positive value indicates the spontaneous nature of the reaction under standard conditions. Calculating this helps us understand the feasibility and spontaneity of electrochemical cells.

Reaction Quotient

The Reaction Quotient, \(Q\), measures the relative amounts of products and reactants during a reaction at any point in time. It has the same form as the equilibrium constant, \(K\), but differs in that it applies to non-equilibrium states.\
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In the electrochemical context, the reaction quotient can be expressed with concentrations of gases and ions involved in the reaction:
\[Q = \frac{[\mathrm{H}_2\mathrm{O}]^2[\mathrm{Br}_2]^2}{[\mathrm{O}_2][\mathrm{H}^+]^4[\mathrm{Br}^-]^4}\]However, because \([\mathrm{H}_2\mathrm{O}]\) and \([\mathrm{Br}_2]\) are typically constant in liquid form, they simplify in calculative expressions. Thus, \(Q\) is inversely proportional to concentrations in the denominator. In our exercise, \([\mathrm{O}_2]\), \([\mathrm{H}^+]\), and \([\mathrm{Br}^-]\) affect \(Q\).
Knowing \(Q\) allows us to determine how far a reaction has progressed and at what potential it operates. Using the Nernst Equation, \(Q\) is used to adjust the standard potential mode to reflect non-standard operational states.